Let a, b, and c be such that frac{1}{(1-x)(1- | vedclass

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(A) Given the partial fraction decomposition: $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$.Multiplying both sides b

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$\frac{1}{15}$

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(A) Given the partial fraction decomposition: $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$.Multiplying both sides by $(1-x)(1-2x)(1-3x)$, we get:$1 = a(1-2x)(1-3x) + b(1-x)(1-3x) + c(1-x)(1-2x)$.To find $a$, set $x = 1$: $1 = a(1-2)(1-3) = a(-1)(-2) = 2a \Rightarrow a = \frac{1}{2}$.To find $b$, set $x = \frac{1}{2}$: $1 = b(1-\frac{1}{2})(1-\frac{3}{2}) = b(\frac{1}{2})(-\frac{1}{2}) = -\frac{1}{4}b \Rightarrow b = -4$.To find $c$, set $x = \frac{1}{3}$: $1 = c(1-\frac{1}{3})(1-\frac{2}{3}) = c(\frac{2}{3})(\frac{1}{3}) = \frac{2}{9}c \Rightarrow c = \frac{9}{2}$.Now, calculate $\frac{a}{1} + \frac{b}{3} + \frac{c}{5} = \frac{1/2}{1} + \frac{-4}{3} + \frac{9/2}{5} = \frac{1}{2} - \frac{4}{3} + \frac{9}{10}$.Finding a common denominator of $30$: $\frac{15}{30} - \frac{40}{30} + \frac{27}{30} = \frac{15 - 40 + 27}{30} = \frac{2}{30} = \frac{1}{15}$.

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

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